Step 1: Understanding the Concept:
A series resistor \(R_S\) is used to drop the excess voltage from the source to prevent the LED from drawing too much current and burning out. The parallel Zener diode in the given circuit diagram is reverse-biased and does not conduct below its breakdown, meaning all current flows through the LED branch.
Step 2: Key Formula or Approach:
LED parameters: \(P = V_{\text{LED}} \times I\)
Voltage loop (KVL): \(V_{\text{source}} = V_{R_s} + V_{\text{LED}}\)
Ohm's Law: \(R_S = \frac{V_{R_s}}{I}\)
Step 3: Detailed Explanation:
Calculate the operating voltage of the LED:
\[ V_{\text{LED}} = \frac{P}{I} = \frac{2}{0.5} = 4 \, \text{V} \]
The source voltage is \(5 \, \text{V}\). The series resistor \(R_S\) must drop the remainder of the voltage:
\[ V_{R_s} = V_{\text{source}} - V_{\text{LED}} = 5 \, \text{V} - 4 \, \text{V} = 1 \, \text{V} \]
Calculate the required resistance using Ohm's Law:
\[ R_S = \frac{V_{R_s}}{I} = \frac{1 \, \text{V}}{0.5 \, \text{A}} = 2 \, \Omega \]
Step 4: Final Answer:
The minimum resistance is \(2\Omega\).
